HW2_Sol - Math 104 Fall 2009 Homework 2 solution set We use...

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Math 104 Fall 2009 Homework 2: solution set We use the notations below to ease readability. Matrices are bold capital, vectors are bold lowercase and scalars or entries are not bold. For instance, A is a matrix and a ij (sometimes, A ( i, j )) its ( i, j )th entry. Likewise x is a vector and x j its j th component. The linear span generated by a group of vecotors v 1 , v 2 , . . . , v n is denoted by span( v 1 , v 2 , ..., v n ). Problem 1 Since ( AB )( B - 1 A - 1 ) = A ( BB - 1 ) A - 1 = AIA - 1 = AA - 1 = I , we have that AB is invertible and the inverse of AB is B - 1 A - 1 . Problem 2 Suppose that R = r 11 · · · r 1 m . . . . . . . . . r m 1 · · · r mm = [ r 1 , · · · , r m ] is a nonsingular upper-triangular matrix, which means { r 1 , · · · , r m } forms a basis of C m , and r ij = 0 for i > j . Suppose e j is the canonical unit vector with 1 in the j th entry and zeros elsewhere. Then r j = j i =1 r ij e i , which implies r j span( e 1 , . . . , e j ) and so span( r 1 , . . . , r j ) span( e 1 , . . . , e j ). Since r 1 , . . . , r j are linearly independent, we have dim(span( r 1 , . . . , r j )) = j = dim(span( e 1 , . . . , e j )). Thus we
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